Question

If $y^2=P\left(x\right)$ is a polynomial of degree 3, then $2\frac{d}{dx}\left[y^3\frac{d^{2}y}{d{x}^2}\right]$ equals

• A. P’’’ (x) + P’ x
• B. P’’ (x) · P’’’ (x)
• C. P(x) · P’’’ (x)
• D. None of these

Answer 1

The correct option is C P(x) · P’’’ (x)

$y^2=P\left(x\right)\Rightarrow 2y{y}^{\prime }=P^{\prime }\left(x\right)\Rightarrow \left(2y\right)y^{\prime \prime }+y^{\prime }\left(2y^{\prime }\right)=P^{\prime \prime }\left(x\right)\Rightarrow 2y{y}^{\prime \prime }=P^{\prime \prime }\left(x\right)-2\left(y^{\prime }\right)2\Rightarrow 2y3y^{\prime \prime }=y2P^{\prime \prime }\left(x\right)-2\left(y{y}^{\prime }\right)2=y^2{P}^{\prime \prime }\left(x\right)-2\frac{\left(p^{\prime }\right(x){)}^2}{4}[fromEq.(i\left)\right]\Rightarrow 2y^3{y}^{\prime \prime }=P\left(x\right)P^{\prime \prime }\left(x\right)-\frac{1}{2}\left\{p^{\prime }\right(x){\}}^2\Rightarrow 2\frac{d}{dx}(y^3\frac{d^{2}y}{d{x}^2})=P(x)P""(x)$